∵a²+2a-1=0
∴a²+2a=1
(a-2)/(a²+2a)-(a-1)/(a²+4a+4)-(a-4)/(a+2)....[猜着修改的]
=(a-2)/(a²+2a)-(a-1)/(a²+4a+4)-(a-4)/(a+2)
=(a-2)/a(a+2)-(a-1)/(a+2)²-(a-4)/(a+2)
=(a²-4)/a(a+2)²-a(a-1)/a(a+2)²-(a-4)/(a+2)
=(a-4)/a(a+2)²-(a-4)/(a+2)
=(a-4)[1/a(a+2)²-(a²+2a)/a(a+2)²]
=(a-4)[1/a(a+2)²-1/a(a+2)²]
=(a-4)×0
=0
a2+2a-1=0
=> (a2+2a+1)-1-1=(a+1)2-2=0
=> (a+1)2=2
=>a=√2-1
由于题目中的式子写的有歧义，你自己把a值代入要求式子即可求的结果。
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